# Why does a jumping rope have no bending stiffness and compression stiffness?

Can anyone just tell me that why doesn't a jumping rope, or any other kind of rope, have any bending stiffness? If I hold one end of the rope in my hand, the rest of the rope will just fall. It won't just keep itself straight. I know it is because of weight of the rope, but why can't a rope carry a bending moment at all? It is very strong in the axial direction, but it can't carry any compression load or any bending loads.

I mean what parameters determines that these structures would be able to resist bending moments but these won't? And why can't it carry compression loads as well? A detailed answer would be extremely appreciated.

• ropes do have some bending stiffness. If they had zero, the would drop 90 degrees from the handle. A chain will have zero when taken across links. Dec 29, 2021 at 20:04
• because a 1cm-thick rope is not a 1cm thick thing being bent. it is a multitude of microscopic threads being bent. length to thickness aspect of the threads in a rope is millions to one, and nothing at that aspect ratio is strong at resisting bending, or linear compression (due to buckling) Dec 31, 2021 at 8:42

TL;DR: The bending and compression (buckling) stiffness is so small because the second moment of area of the fibres is small.

## Bending stiffness

It does have a bending stiffness however it is really really small. More precisely it has a really really small second moment of area per fibre. (see the numerical example below to understand why many more independent fiber don't perform as well.

And since every fibre is separate from the others fibres only the thickness of each fibre is contributing to the second moment of area of a fibre. The second moment of area is :

$$I = \frac{\pi D^4}{64}$$

The bending stiffness (or more precisely one of the measures for it) is a product of the Young's Modulus and second moment of area.

## Compression Stiffness

Actually this is not Compression stiffness but its Buckling related stiffness. Again, the critical load in (Euler) buckling is a function of the second moment of area:

$$P_{crit} = \left(\frac{\pi}{L^2}\right)EI$$

where :

• $$L$$ is the critical lenght
• $$E$$ is the young's modulus
• I is the second moment of area.

## A numerical example

assume that you have a fiber with a square cross-section of a= 1mm(this is to simplify the comparison because for what I want to do below it doesn't scale that well). Then the equation of the second moment of area would be slightly different to the circular cross-section and equal to $$I_{1f} = \frac{a\cdot a^3}{12}= \frac{1}{12} mm^4$$

If you stacked (in line 1000 of those fibers without gluing them together) then the total stiffness would be the sum of all the fibres:

$$I_{1000f} = 1000 \cdot I_{1} = \frac{1000}{12} mm^4$$

However, if all fibres were bonded together along the direction of the applied load (see image below)

then the total second moment of area would be equal to:

$$I_{bonded} = \frac{a \cdot (1000\cdot a)^3}{12} = 10^9 \frac{a \cdot (a)^3}{12} = 10^9 I_{1f} = 10^6 I_{1000f}$$

Hopefully, it is obvious why this makes a huge difference. Of course the fibre in the yarn doesn't behave entirely independently from the other fibres in the yarn, like the example I provided above (and also the fibres in other strand have less dependence compared to the fibres in the same yarn). So, all in all, for explaining purposes IMHO the independence simplification is a fair compromise.

Essentially this is very similar to the bending and compression stiffness of a book -- each page is a flat fibre and it can move independent of the adjacent. If you somehow made the pages glued together then the thickness would increase manyfold.

## Another example

The following is an easy to perform real life example that everybody can try. I took a simple paperback book and

1. held it in my hand like a cantilever
2. applied a clothespin at the edge of the book to couple the motion of each page with its adjacent.
Image Description
independent
with clothespin

Although:

• the pages are not bonded and the constraint applies only at the end of the pages
• the clothespin adds a weight at the end (and more significantly more bending moment) the deflection is visibly smaller in the case of the clothespin (you can try this at home).
• I couldn't comprehend why you said that a simple rope (with all the fibers) would have moment of inertia equal to the sum of all the fibers. Look, if the cross section has like 1000 fibers in a line, then the overall moment of inertia would be equal to sum of moment of inertia of a fiber about its own neutral axis + (perpendicular distance from centroid of this fiber to centroid of the overall rope's cross sectional centroid * area of this fiber). However, you didn't include the latter term. Why though? Dec 29, 2021 at 19:42
• So basically, what you mean is that we cannot use the parallel axis theorem when each fiber is allowed to behave independently, without affecting the behavior of rest of the fibers. Did I get that right? Dec 29, 2021 at 20:28
• A note on what 'every fibre is separate from the others fibres' means: in analysing the ability of a beam to bear a bending moment, the shear tractions on planes perpendicular to the length of the beam are crucial... Dec 30, 2021 at 14:50
• ... But if there are shear tractions on those planes, the mechanical requirement that shear stress is a symmetric tensor means there will also be shear tractions on some of the contact surfaces between individual fibres... Dec 30, 2021 at 14:52
• @RameezUlHaq In the Wikipedia article I linked before, there's a list of seven assumptions of Euler-Bernoulli beam theory. The inability of the contact surfaces between fibres in a rope to bear as much shear stress as intact material violates assumptions 2, 3, and 7. Dec 30, 2021 at 16:44

Having low stiffness is part of the specification of a rope. It would be far cheaper to obtain the same tensile strength with a solid rod. Creating and manipulating multiple strands is expensive.

It is designed to use multiple thin strands to achieve this lack of stiffness.

The question of how thin strands, with their small moment of area, achieve this lack of stiffness is well covered by the other answers.

It won't just keep itself straight

You know a steel beam held at one end won't just keep itself straight either right? It also bends under it's own weight when supported by one end. Just because you can't perceive something with eyes doesn't mean it's not there. It just means your eyes aren't precise enough.

The fact that the rope has a radius of curvature when you hold one end of it as it bends under its own weight indicates it does have some compressive strength and stiffness and can carry a bending moment. And if you've ever held a steel wire rope, as some jump ropes are, they definitely aren't very floppy and have a very noticeable radius of curvature.

Engineering point of view:

There are ropes that do have significant bending stiffness (and, as a consequence, compression hardness). E.g. steel wire ropes.

This is almost never an advantage - such ropes tend to self-tangle easier (but making intentional knots is harder), tend to break at small bending radius (making most knots unusable) and tend to retain their bent shape after winding/unwinding.

One uses such ropes when the alternatives get impractically thick or impractically expensive.

They design the jumping rope this way. if it would have any moment strength or compression strength it woul act as a soft "column" at the strat where half of the cord is bent up to reach the Jumper's harness.

Thus it could send the jumper into an unpredictable dangerous arc, as opposed to let them fall straight down into hopefully a clear vertical path.

## Edit

After reading some comments I realized my mistake. I can assign some of the blame to the use of my wife's reading glasses after losing mine in a ski crash. I thought it was Bungee jumping rope, apologies. Please accept that answer for a bungee rope.

• A rope with no moment or axial bearing capacity will assume a catenary shape and will not have a tendency to vibrate on its own, $$\omega=\sqrt{\frac{k}{m}}, \quad if,\ k=0\ \omega=0$$ making handling hard.
• A rope capable of supporting only tensin will be very easy to control and feel its feedback because the tension transferred to the hands is,$$F=1/2 (m_{rope}\pm \text{centripetal force of the catenary}$$, but a rope with stiffness would introduce confusing forces due to vibration.