Thin wire is springier than thick wire?

Does thin spring type wire flex more than thick spring type wire (everything else equal), before losing its shape?

Putting several thin wires together will provide the same amount of strength as a thick wire, with more flexibility?

Reply to
John Doe
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Some handguns have slide springs made this way- out of several wires twisted together, then wound into a spring.

It's just a guess, but durability and reliability may be a factor here.

Reply to
Cydrome Leader

Yes. But it has less resistance to flex.

No.

Nick

Reply to
Nick Mueller

Indeed. My daily carry has just such a spring. Hasnt lost any significant power in 10 yrs and 20,000 rds

When you care enough to carry the very best.....

Gunner

A human being should be able to change a diaper, plan an invasion, butcher a hog, conn a ship, design a building, write a sonnet, balance accounts, build a wall, set a bone, comfort the dying, take orders, give orders, cooperate, act alone, solve equations, analyze a new problem, pitch manure, program a computer, cook a tasty meal, fight efficiently, die gallantly. Specialization is for insects. Lazarus Long

Reply to
Gunner

Agreed, if "flex more...before losing its shape" is interpreted as, "you can bend it farther before it will permanently deform."

I was going to say yes, but I'm taking "same amount of strength" to mean equal tensile, and "more flexibility" as lower spring rate.

Which goes to show the importance of using technically correct, or at least unambiguous, terms when framing such a question.

Reply to
Ned Simmons

Say: It can be bent in a smaller radius (or compressed more) before it is permanently bent.

Same spring rate. The answer is "No".

Here, the answer is "Yes". :-)

Nick

Reply to
Nick Mueller

Maybe I'm wrong with my "No". Let's see: If I have a thin wire, it can be bent much more like a thick one (think glass fibers; even glass can be bent). Now if I put a lot of springs made out of thin wire side by side (or arrange it in some other way), I can get the same spring rate like the thick-wired spring, but it can be bent more before it fails.

Now what's wrong with *that*?

Nick

Reply to
Nick Mueller

Yes, a precise literal translation of what I wrote might sound impossible.

The tensile strength of several small wires would make the same tensile strength of one large diameter wire.

Flex means "to bend especially repeatedly". I believe that fits the context very well.

If you thought I meant the set-in-stone literal translation of "flex before losing its shape", you are creative as a CPU.

Reply to
John Doe

In article , John Doe writes

A very confusing discussion so far, with a lot of vague terminology and misleading statements (with the honourable exception of Nick).

Any rod, when bent, will be stretched on the outside (tensile strain) and compressed on the inside (compressive strain). The amount of the strain, for any bend radius, will depend on the thickness of the wire, and indeed will be proportional to the thickness. Thus a thin wire can be bent to a smaller radius before experiencing the same strain as a thick wire. You can easily see this by drawing, if you compare the inner/outer arc lengths of a thin annulus and a thick one.

At some level of strain, the outer layers of the wire will reach its yield stress (and remember stress = strain times Young's modulus, which is a constant for a given metal) and will take on a permanent set or stretch - it will not go back to its original shape when the stress is removed.

Since the thin wire has lower strain, and stress, levels at a given bend radius, it will go to a lower radius before it takes on a set.

The force required to make a particular bend is a whole different issue. Assuming this is a cantilever spring (i.e. a straight wire held at one end and having lateral force applied at the other end) the deflection will be proportional to 1/d^4, where d is the diameter of the wire*.

Thus if a wire of thickness 1.0 deflects by D under a force F, a wire of thickness 0.5 will only require a force of F/16 to deflect the same distance D, but it can deflect much further (to half the radius) before it takes a permanent deformation.

You can then see that the answer to your second question is, yes, but to get the same "resistive force" (spring rate) you need 16 wires of (say)

0.5mm diameter to replace one wire of 1.0mm diameter. It will bend further but take up a lot more room. Whether this is reasonable is up to you. *The actual equation, for circular wire, is D = 64FL^3/(3piEd^4)

HTH

David

Reply to
David Littlewood

I'd say that pompous judgment comes from your dislike for Americans and the fact Nick is the only other obviously non-American in this thread.

By the way, and for what it's worth. According to the International Aluminium Institute, the word "aluminum" came before "aluminium". It wasn't "Bloody Americans who can't spell getting their nasty bastardised spelling of aluminum adopted".

Wow.

Yeah, a non-scientific mind might think about Nick's example of glass. One can imagine many flexible glass strands would be needed to produce the hardness of a common piece of glass window.

Anyway. Thank you (sincerely) for the technical explanations, oh most honorable David Littlewoody.

Reply to
John Doe

Well, I thought I was helping you. If I had known what an ungrateful bastard you were, I would have saved my time.

Reply to
David Littlewood

...

I thought you were being a pompous anti-American ass.

Okay Littlewoody.

Reply to
John Doe

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