Deflection of an inclined rope

[kalleK]

Hi,

I am looking for a formula that describes the deflection of an inclined rope that is fixed in both ends, the left end is at the origin and the right end is at (xB, yB). The known parameters are; the length between the ends is L, the length of the rope is l, the initial angle is β and there is a uniform force, Q, acting on the rope.

I have tried to use ∑Fx = 0, ∑Fy = 0 and ∑MA = 0 but i can't find a good combination that describes the deflection of the rope.

A and B corresponds to the left and right end respectively.

 

[Orodruin]

https://en.wikipedia.org/wiki/Catenary

 

[kalleK]

https://en.wikipedia.org/wiki/Catenary

I have already tried to look at this problem as a catenary, but the tension force and the acting force on the rope cancels out which leads to that the deformation of the rope will be same regardless the force on the rope.

 

[Orodruin]

I have already tried to look at this problem as a catenary, but the tension force and the acting force on the rope cancels out which leads to that the deformation of the rope will be same regardless of the force on the rope.

In which variable is the force uniform? The rope length or the $x$ coordinate?

 

[Nidum]

Draw me a picture with a realistic representation of the geometry and loading system relating to this problem .

 

[kalleK]

Don't know how to upload the figure here, but click on the link and the figure will come up
https://ibb.co/ky6uPQ

 

[Orodruin]

Again, it is not clear from that image whether or not the load $Q$ is distributed evenly in the distance along the rope or in the horizontal distance. This is a crucial point as the problem and its solution will be different.

 

[Orodruin]

I have already tried to look at this problem as a catenary, but the tension force and the acting force on the rope cancels out which leads to that the deformation of the rope will be same regardless the force on the rope.

The point is that if the load is distributed evenly in the distance along the rope, then the shape will be the same and independent of the load (as long as the load is not negative and larger than the gravitational force on the rope). You can also see this in the catenary equation, which does not depend on the gravitational acceleration $g$ (changing the gravitational acceleration is equivalent to changing the load on the rope evenly in the distance along the rope). The only thing that will change is the tension in the rope, not the rope's shape.

 

[kalleK]

The point is that if the load is distributed evenly in the distance along the rope, then the shape will be the same and independent of the load (as long as the load is not negative and larger than the gravitational force on the rope). You can also see this in the catenary equation, which does not depend on the gravitational acceleration $g$ (changing the gravitational acceleration is equivalent to changing the load on the rope evenly in the distance along the rope). The only thing that will change is the tension in the rope, not the rope's shape.

A new figure is in the link with the correct $Q$.
Thank you for the explanation. But i have hard to see that the shape is independent of the load, it feels like that the deflection should depend on the load?

 

[Burken]

The point is that if the load is distributed evenly in the distance along the rope, then the shape will be the same and independent of the load (as long as the load is not negative and larger than the gravitational force on the rope). You can also see this in the catenary equation, which does not depend on the gravitational acceleration $g$ (changing the gravitational acceleration is equivalent to changing the load on the rope evenly in the distance along the rope). The only thing that will change is the tension in the rope, not the rope's shape.

But how will the deflection be affected if the problem instead has a un-uniformly distributed load? As KalleK mentioned, the load is canceled out in the function that describes the deflection of the rope. Let's say that the load for a uniformly distributed load is Q = q. If the load instead is un-uniformly distributed, with perhaps a load function described like
Q = q₁*x +...

The load term will still cancel itself out from the deflection term or doesn't the catenary function apply to an un-uniformly distributed load on a rope?

 

[Orodruin]

Thank you for the explanation. But i have hard to see that the shape is independent of the load, it feels like that the deflection should depend on the load?

Why? A catenary on the Moon looks just like it does on the Earth. All forces involved are the load and forces that are reactions to it and grow linearly with it. Of course, things change if you have an elastic rope which needs to deform to change its tension.

But how will the deflection be affected if the problem instead has a un-uniformly distributed load? As KalleK mentioned, the load is canceled out in the function that describes the deflection of the rope. Let's say that the load for a uniformly distributed load is Q = q. If the load instead is un-uniformly distributed, with perhaps a load function described like
Q = q₁*x +...

The load term will still cancel itself out from the deflection term or doesn't the catenary function apply to an un-uniformly distributed load on a rope?

For an arbitrary load you will have to solve the appropriate differential equation anew. The easiest way of obtaining the differential equation is not through force and torque analysis, but through using variational calculus to find the minimum of the potential energy.

 

[Burken]

Why? A catenary on the Moon looks just like it does on the Earth. All forces involved are the load and forces that are reactions to it and grow linearly with it. Of course, things change if you have an elastic rope which needs to deform to change its tension.For an arbitrary load you will have to solve the appropriate differential equation anew. The easiest way of obtaining the differential equation is not through force and torque analysis, but through using variational calculus to find the minimum of the potential energy.

Ok, thank you sir!

 

[kalleK]

Why? A catenary on the Moon looks just like it does on the Earth. All forces involved are the load and forces that are reactions to it and grow linearly with it. Of course, things change if you have an elastic rope which needs to deform to change its tension.

I see, thank you!

 

[Orodruin]

Just a different way of looking at it:
In the case of an inelastic rope in a gravitational field, the rope will take the shape that minimises the potential energy (subject to the constraint that the length of the rope is fixed). In the expression for the potential energy, the gravitational constant is just a multiplicative factor which means that changing it from $g_1$ to $g_2$ just rescales the potential energy as $V \to g_2 V/g_1$. If you multiply any function by a (non-zero and positive) constant, it will not change where the function has its minima, only what the value at the minima are. As a result, the potential energy changes, but the shape remains the same.

 

[kalleK]

Ok, that made it more clear. But what if the rop is elastic, which approach can be used to solve that problem?

 

[Orodruin]

You can use a similar approach with the only difference that you will have to include the rope’s potential due to being stretched instead of the condition of a fixed rope length. However, it will be significantly more difficult to solve the resulting differential equation.

 

[kalleK]

Okay, can you recommend any book for this problem?

 

[Burken]

I have been thinking of this problem. The functions used to calculate the sag and the natural hanging of the rope when load is applied upon it is part of the general cable theorem. We have used certain functions from the book "Statics - Learning from engineering examples" by Igor Emry & Arkady Voloshin.

Please see the link for the function: (The one that is named y)

https://ibb.co/kC7gTkHowever, this function is for a cable/wire, and for our problem, the cable is instead a rope that will have a high elasticity. Much higher than for a cable. Is it still possible to use the same function even though the rope has the much higher elasticity?

/Johan

 

[Nidum]

Are your questions related to obtaining theoretical equations for some academic purpose or to solving a practical problem ? .

 

[Burken]

Are your questions related to obtaining theoretical equations for some academic purpose or to solving a practical problem ? .

It is regarding to obtain theoretical equations for the case of an elastic rope instead of a cable/wire.

 

[Nidum]

I asked because solutions for any specific configuration of this problem could be obtained using direct numerical methods or FEA . It would not actually be necessary to derive the controlling equations in full .

Anyway I don't think that it should be too difficult to derive the theoretical equations from basics . Try solving these three problems sequentially :

A weightless elastic rope set initially at 45 degrees to the vertical and carrying a single load acting vertically downwards at the mid length of the rope . This configuration can be solved quite easily .

A rope set at 45 degrees to the vertical but with three identical and evenly spaced vertical loads and including the weight of the rope . Weight of rope to be considered as discrete point loads acting at same locations as applied loads .

A rope set at any angle and with fully distributed vertical loads and weight of rope .

In each case rope is considered to be just lightly tensioned between supports so that it doesn't have much initial sag .

 

[Nidum]

Another five minute FEA . 3 metre long rubber rope at 45 degrees to vertical . Uniformly distributed load acting vertically downwards . Linear model and magnified deflection for display .

As far as I can tell by visual inspection that curve is symmetrical . I'll have a look at the raw data later on to see if it actually is .

 

[Burken]

View attachment 211710

Another five minute FEA . 3 metre long rubber rope at 45 degrees to vertical . Uniformly distributed load acting vertically downwards . Linear model and magnified deflection for display .

As far as I can tell by visual inspection that curve is symmetrical . I'll have a look at the raw data later on to see if it actually is .

Interesting to see the deflection simulated. Thanks for the input. Knowing the behavior will help us al ot when solving it numerically,