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I was wondering how one would go about figuring out the energy stored in a spring in a clothespin. When opened the spring is deformed, but I don't have any intuition for why it gets stronger with more loops, how it works, or how to calculate the energy or force stored.

Some searching suggests it is a torsion spring, I'll read on that now. Any conceptual help would be appreciated, I find it difficult to understand why so many loops are needed.

Here is the spring:

clothespin spring

And it acts as the hinge like so:

clothespin with spring

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I don't think all of the loops are required. I think it has a bunch of loops because the spring also acts as the hinge. Also, the more loops the less the circumference is increased and the less the spring deforms. –  Brandon Enright May 30 '13 at 5:31
Is that accounted for in the constant? Having no loops, the spring would just bend and deform almost completely-Would infinite loops not bend at all? Why does having more loops decrease the amount the two side pieces of metal bend? –  Anthony May 30 '13 at 5:36
Having bent these sorts of springs way beyond their designed limits, I've noticed they tend to twist rather than continue to bend. I suspect more loops spreads out the bending over more length of wire and reduces the tendency to twist rather than bend. –  Brandon Enright May 30 '13 at 5:46

2 Answers 2

up vote 4 down vote accepted

Yes, it is a torsion spring. It works by twisting the metal rod that makes up the body of the spring.

The reason for coiling the spring is to fit a long length of metal rod into a short space. You need a long length of rod so that the torsion per unit length remains small. With a shorter length of rod you'd exceed the elastic limit and the rod would be permanently twisted.

Don't be deceived that the body looks like a conventional spring. The only reason for coiling the rod into a helix is to fit a long length of rod into a short space and although it may look like a conventional coiled spring that is not how it works.

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I thought the twist was a failure mode. I didn't realize that twisting was actually how the spring worked. Very cool, thanks! –  Brandon Enright May 30 '13 at 7:36
@BrandonEnright: Actually, conventional springs also work by twisting the wire they are made from. We just don't call them torsion springs because a torsion spring normally refers to a straight length of metal. Topologically, coil springs, clothes peg springs and torsion springs all work in the same way. –  John Rennie May 30 '13 at 7:39

I would use the well known formula of

$$ U = \int \frac{T^2 }{2 G J} + \frac{F^2 }{2 A G}\;{\rm d}l $$

where $T$ is the torque on the coils, $F$ is the axial force, $A$ is cross section area, $G$ is the modulus of rigidity and $J$ is the polar moment of area. The axial length $l$ is only that if the active coils that counts.

If the applied force $F$ is a distance $a$ from the coil center, then the internal torque in the coils is

$$ T = F a + F \frac{D}{2} \sin\theta $$

where $D$ is the spring diameter and $\theta$ the location along the coil, starting with $\theta=0$. If the active coils are $N$ then the total angle along the helix is $\Theta = N 2 \pi$ and the length increment is ${\rm d} l = \frac{l}{2\pi N} {\rm d}\theta$. Also for a round wire $A=\pi\frac{d^2}{4}$ and $J=\pi\frac{d^4}{64}$, with $d$ the wire diameter.

The total elastic energy is

$$ U = \frac{4 F^2 l ( D^2+8 a^2)}{\pi G d^4} + \frac{2 F^2 l}{\pi G d^2} $$

for integer number of coils only, such that $\cos(2\pi N)=1$.

To get the deflection at the point of force $F$ you calculate $\delta = \frac{\partial U}{\partial F}$.

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