- #1
phantomvommand
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- Homework Statement
- See picture below
- Relevant Equations
- F = kx
Resolving forces?
Is this even solvable? I don't know where to begin.
Tension in a spring that's bent into a semicircle
In summary, you are suggesting that the tension in the string arises from the fact that the inner circumference of the spring is greater than the distance around its outer perimeter.
Is this even solvable? I don't know where to begin.
- #2
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Judging from the diagram, the spring starts compact, i.e. it cannot be compressed (much).phantomvommand said:
Consider the outermost parts of the spring in the curved position. What length is that now?
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- #3
phantomvommand
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Are you suggesting that the diameter of the semicircle is (D + 2L/pi)?haruspex said:
I think I need help with understanding how a vertical tension in the string even arises. It is obvious intuitively, but when I break the spring down into an infinitesimal element I cannot figure out why.
Considering the small element of the spring where the string is attached to, it experiences a leftward spring / tension force, and a downward tension force from the string. Yet, it is in equilibrium. Presumably, the leftward tension is balanced by some rightward friction with the string? But how about the downward tension force? What gives rise to a necessary upward force which the downward string tension must balance?
If my method of analysis above is incorrect, I'd be happy to hear how else the tension in the string arises.
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Yes, it is non obvious.phantomvommand said:
When a helical spring is stretched, each small cylindrical element of the wire is twisted about its axis. So it involves shear moments, like a torsion wire. (Confusingly, a "torsion spring" works through bending moments.)
However, you are probably not expected to get into such detail.
Nearly right… think that through again.phantomvommand said:
But it is not the diameter that is of direct relevance; as I asked, what is the distance around the outer perimeter of the curved spring? Hence, how much has the spring been stretched on that side?
- #5
Lnewqban
Homework Helper
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The thing is, as the picture shows, that bent spring will not take the shape of a semicircle.phantomvommand said:
The apex is under a greater amount of bending moment than the cross-sections closer to the ends.
What is the subject to which this problem is related?
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True, but the question explicitly states it is to be taken to be a semicircle.Lnewqban said:
- #7
phantomvommand
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If I am understanding correctly, you are suggesting that assuming the spring is not stretched much, it's inner circumference is L, which gives an outer perimeter length of (L + D*pi).haruspex said:
Are you suggesting that the tension is then kD*pi?
I am a little confused as to how such a tension translates into a net vertical tension required at the ends with the string, and furthermore, why is this tension, which is calculated only for the outer perimeter of the spring, representative of the overall tension at all points in the spring?
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Not quite. As you note, that is only for the most stretched arc through the spring. What would it be on average? Bear in mind the circular nature of each coil of the spring.phantomvommand said:
The next step would be to consider moments.phantomvommand said:
Having found the effective tension in the spring, think of the spring as two rigid quadrants hinged by a single coil of the spring. Take the balance of moments about the innermost point of that coil.
- #9
Lnewqban
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A spring does not behave like a rubber band.phantomvommand said:
Besides accepting stretching, it can be compressed (if pitch of coils allows space for it), and can also be bent, or even twisted.
Our spring has been bent until reaching the shown shape.
As I can see spaces between the coils still in the bent position, I would assume that there is a neutral axis, a compressed concave side and a stretched convex side (just like it happens to a beam subjected to bending loads).
Please, see:
https://en.wikipedia.org/wiki/Bending
If that is true, at each end of the spring, a moment had to be applied, in order to achieve a semi-circular shape.
The magnitude of that moment should be equal to stretching force times D, or compressing force times D, or (stretching force + compressing force) times D/2.
If you now attach the string to both ends, and remove the bending moments applied to the ends, the spring can’t keep the semicircular shape, but a tension appears in the string.
There is now an internal moment that grows in magnitude from each end to the apex, where is value is maximum.
The tension of your string is now perfectly balancing all those internal moments, since the spring is in repose.
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As I wrote in post #2, it doesn’t look that way to me (except near the ends, where the moment from the string is weak).Lnewqban said:
A moment had to be applied whether or not the coils are touching their neighbours on the inside of the curve.Lnewqban said:
Lnewqban said:
I don't understand those last two points. The sequence you described seems to beLnewqban said:
- manually bending the spring into the arc
- attaching the string
- releasing the spring so that the string is in tension
- #11
Lnewqban
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My post was directed to the OP, who stated he needs help understanding the reason for the string tension (please, see selected header quote in my last post).haruspex said:
As you may know by now, sometimes I induce confusion as I try to explain my ideas, as English is not my native language.
My apologies, @haruspex
FAQ: Tension in a spring that's bent into a semicircle
What is the formula for calculating the tension in a spring bent into a semicircle?
The tension in a spring bent into a semicircle can be calculated using the formula T = (k * ΔL) / (2 * π * r), where k is the spring constant, ΔL is the change in length of the spring, and r is the radius of the semicircle.
How does the curvature of the spring affect the tension?
The curvature of the spring affects the distribution of tension along its length. In a semicircular spring, the tension is not uniform but varies along the arc. The maximum tension typically occurs at the ends of the semicircle.
What role does the spring constant play in determining the tension?
The spring constant (k) is a measure of the stiffness of the spring. A higher spring constant means the spring is stiffer and will exert more force for a given deformation, resulting in higher tension when the spring is bent into a semicircle.
How do you account for the length of the spring in the tension calculation?
The length of the spring (L) affects the change in length (ΔL) when the spring is bent. For a spring bent into a semicircle, the length of the spring is half the circumference of the circle, L = π * r. This length is used to determine the deformation and, consequently, the tension.
Can external forces or constraints alter the tension in a semicircular spring?
Yes, external forces or constraints can significantly alter the tension in a semicircular spring. External forces, such as applied loads or fixed supports, can change the distribution and magnitude of tension along the spring. Constraints that limit the spring's movement can also affect how the tension is distributed.