# Which steel rod would have the least sag/twist/flex of a 4' stretch

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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?

Jeremy

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

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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.

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correct.

Fairly understandable essay on deflection of beams.

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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

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With no more than 0.0003" sag, right?

Maybe something like this is in your future:

Support that puppy!

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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

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Can you explain why it is so?

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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

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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

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

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"Carl Ijames" fired this volley in news: snipped-for-privacy@news4.newsguy.com:

Bravo, Carl! A good, short treatise!

Lloyd

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What Carl said. If you mean what is the physics behind it, no, you'd have to look that up.

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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.

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

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The Wiki describes elastic behavior in more detail, with examples:

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.

'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

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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.

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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

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Thanks again everyone, the answers are all really appreciated, and I think I understand, but just in case;

If I understand what has been said it does not matter if it is 1144 Stressproof or 4140, it is going to sag under its own weight if it is 3/8" X 4'. The treatment to the steel or the carbon percentage is only going to affect how far it can bend or twist while still being able to return to its original shape or before it will snap. Am I correct in my thinking on this? I haven't checked out all of the links that were provided, but I guess I could find a formula to see how much the sag will be at the center?

Would it be the same with a chromium vanadium steel alloy, or something like that?

Thanks again everyone

ATTN ADMIN: I posted the first message under Jman before I created this account, so I didnt know if you wanted to change the author on that post. Great site

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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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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

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

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Coat hangers are also good for replacing 9 wire as drop-in light supports to Hilti straps (for drop ceilings).

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Like Ed said, the type and hardness has virtually no difference in stiffness. So any steel of any hardness will sag the same amount. I know, it seems like that can't possibly be true because heat treated steel is used all the time when mild steel won't work because it is too weak. But as long as the elastic limit is not reached all steels will deflect almost exactly the same amount when the same amount of force is put on these steels. This doesn't mean the steel has to snap, as you put it, it just means permamanently deformed, in other words the steel stays at least a little bent when the elastic limit is surpassed. Tensile strength and yield strength will of course vary tremendously from the weakest to the strongest steels while stiffness will only vary a tiny amount. Go figger. Eric

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